The electric potential at distances of 2.00 m, 4.00 m, and 12.0 m from the line charge are approximately 1.35 * 10^6 V, 6.74 * 10^5 V, and 2.24 * 10^5 V, respectively, when referenced to a distance of 2.50 m from the line of charge where V = 0.To find the electric potential at different distances from the infinite line charge, we can use the formula for the electric potential due to a line charge:
V = kλ / r
where V is the electric potential, k is the electrostatic constant (8.99 * 10^9 N·m²/C²), λ is the linear charge density (in C/m), and r is the distance from the line charge.
Given that the linear charge density is +1.50 μC/m (1.50 * 10^-6 C/m), we can calculate the electric potential at the given distances:
a) At a distance of 2.00 m:
V₁ = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 2.00 m
b) At a distance of 4.00 m:
V₂ = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 4.00 m
c) At a distance of 12.0 m:
V₃ = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 12.0 m
Now, to calculate the potential with respect to a reference point at a distance of 2.50 m from the line of charge, we subtract the potential at that reference point from each of the calculated potentials:
V₁' = V₁ - V(2.50 m)
V₂' = V₂ - V(2.50 m)
V₃' = V₃ - V(2.50 m)
Given that V(2.50 m) = 0 (as chosen in the problem), the equations simplify to:
V₁' = V₁
V₂' = V₂
V₃' = V₃
Now, we can substitute the known values and calculate the electric potential at each distance:
a) At a distance of 2.00 m:
V₁' = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 2.00 m
b) At a distance of 4.00 m:
V₂' = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 4.00 m
c) At a distance of 12.0 m:
V₃' = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 12.0 m
Calculating the results:
a) V₁' ≈ 1.35 * 10^6 V
b) V₂' ≈ 6.74 * 10^5 V
c) V₃' ≈ 2.24 * 10^5 V
Therefore, the electric potential at distances of 2.00 m, 4.00 m, and 12.0 m from the line charge are approximately 1.35 * 10^6 V, 6.74 * 10^5 V, and 2.24 * 10^5 V, respectively, when referenced to a distance of 2.50 m from the line of charge where V = 0.
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What is the net electric field at x=−4.0 cm ? Express your answer using two significant figures. Two point charges lie on the x axis. A charge of 6.0μC is at the origin, and a charge of −9.1μC is at x=10.0 cm You may want to review (Pages 671−675 ) Part B What is the net electric field at x=+4.0 cm ? Express your answer using two significant figures.
To determine the net electric field at a specific point on the x-axis, we need to consider the individual electric fields created by each charge and their respective directions.
Given:
Charge at the origin (x = 0):
q1 = 6.0 μC
Charge at x = 10.0 cm:
q2 = -9.1 μC
Point of interest: x = ±4.0 cm
The electric field at a point due to a point charge can be calculated using the equation:
[tex]E = k * (q / r^2)[/tex]
Where:
E is the electric field
k is the electrostatic constant (9 x 10^9 N m^2/C^2)
q is the charge
r is the distance between the charge and the point
Let's calculate the electric field at x = -4.0 cm:
Distance from q1 to the point (-4.0 cm):
[tex]r1 = 4.0 cm \\= 0.04 m[/tex]
Electric field due to q1:
[tex]E1 = k * (q1 / r1^2)[/tex]
Distance from q2 to the point (-4.0 cm):
[tex]r2 = 10.0 cm + 4.0 cm \\= 14.0 cm \\= 0.14 m[/tex]
Electric field due to q2:
[tex]E2 = k * (q2 / r2^2)[/tex]
The net electric field at x = -4.0 cm is the vector sum of E1 and E2. Since q2 is negative, the electric field due to q2 will have the opposite direction compared to E1.
Net electric field at x = -4.0 cm:
[tex]E_{net} = E1 - E2[/tex]
Now let's calculate the electric field at x = +4.0 cm:
Distance from q1 to the point (+4.0 cm):
r1 = 0.04 m
Electric field due to q1:
[tex]E1 = k * (q1 / r1^2)[/tex]
Distance from q2 to the point (+4.0 cm):
[tex]r2 = 10.0 cm - 4.0 cm \\= 6.0 cm \\= 0.06 m[/tex]
Electric field due to q2:
[tex]E2 = k * (q2 / r2^2)[/tex]
The net electric field at x = +4.0 cm is the vector sum of E1 and E2.
Net electric field at x = +4.0 cm:
[tex]E_{net} = E1 + E2[/tex]
Please note that the value of k is 9 x 10^9 N m^2/C^2.
Now let's calculate the net electric fields at x = -4.0 cm and
x = +4.0 cm using the given charges.
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A moure is eating cheese 309 meters from a sleeping cat. When the cat wakns up, the mouse immediately starts running away from the cat with a constant velocity of 1.29 mis. The cat immediately starts chasing the mouse with a constant velocity of 4.23 m/s. Assume the cat and the mouse start running intantancousy wo there are no accelerations to worry about. Qin vour answers to 2 dncemal places. Wow many wecenis after they begin renning does the cat eateh the mouse? Hew tar does the cat have te run ta catch the mouse? meters A moute is eating cheeve 3.89 meters from a sleeping cat. When the cat wakes up, the mouse immediately starts running away from the cat with a constant velocity of 1.29 m/s. The cat immediately starts chasing the mouse with a constant velocity of 4.23 m/s. Assume the cat and the mouse start running. instantaneously so there are no accelerations to warry about. Give your answers to 2 decimal places. How many seconds after they begin funning does the cat catch the mouse? seconds How far does the cat have to run to catch the mouse? metery
A moure is eating cheese 309 meters from a sleeping cat. When the cat wakns up, the mouse immediately starts running away from the cat with a constant velocity of 1.29 mis. The cat immediately starts chasing the mouse with a constant velocity of 4.23 m/s.The cat catches the mouse approximately 0.92 seconds after they start running. The cat has to run approximately 1.19 meters to catch the mouse.
To solve this problem, we can use the formula:
Time = Distance / Velocity
Let's calculate the time it takes for the cat to catch the mouse:
Time for the mouse to start running away from the cat:
Distance = 309 meters
Velocity = 1.29 m/s
Time = Distance / Velocity
Time = 309 meters / 1.29 m/s
Time ≈ 239.53 seconds
Time for the cat to catch the mouse:
Velocity (cat) = 4.23 m/s
Time = Distance / Velocity
Time = 3.89 meters / 4.23 m/s
Time ≈ 0.92 seconds
Therefore, the cat catches the mouse approximately 0.92 seconds after they start running.
Now, let's calculate the distance the cat has to run to catch the mouse:
Distance = Velocity (mouse) × Time
Distance = 1.29 m/s × 0.92 seconds
Distance ≈ 1.19 meters
Therefore, the cat has to run approximately 1.19 meters to catch the mouse.
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If you charge your céll phone battery with 100 units of energy and you get 120 units of energ) out of the battery, which law of thermodynamics are you violating? 1. A) \( 0^{\text {th }} \) Law
The situation you described, where you charge your cellphone battery with 100 units of energy and get 120 units of energy out of the battery, violates the First Law of Thermodynamics, also known as the Law of Conservation of Energy.
The First Law of Thermodynamics states that energy cannot be created or destroyed within an isolated system. It can only be converted from one form to another or transferred between different parts of the system. In simpler terms, the total amount of energy in a closed system remains constant.
In the case of your cellphone battery, charging it with 100 units of energy and extracting 120 units of energy would imply that the battery is generating energy on its own, which contradicts the principle of conservation of energy. Such a violation would imply the creation of energy from nothing, which is not possible according to our current understanding of physics.
Therefore, the situation you described goes against the First Law of Thermodynamics. In reality, there are losses associated with energy conversions and transfers, such as heat dissipation and inefficiencies in the charging and discharging processes, which would prevent you from obtaining more energy from the battery than you put into it.
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An electron beam is diffracted from the crystal plane (220) of cubic sample with lattice constant 5.43 A at diffraction angle 3.66^ ∘. What is the energy of the incident electrons.
The energy of the incident electrons is 1.213 keV.
Diffraction angle = θ = 3.66°
Lattice constant = a = 5.43 A= 5.43 × 10⁻¹⁰ m
Crystal plane = (220)
For a cubic crystal, the atomic spacing between (hkl) planes is given as
a/hkl = [tex]\sqrt{2}[/tex] / hkl ---(1)
The energy of an electron beam is given as
E = (hc) / λ ---(2)
where
h is Planck's constant, c is the velocity of light, and λ is the wavelength of the electron beam.
The Bragg's law for diffraction is given as
kλ = 2d sinθ ---(3)
where k is a positive integer
Now we need to find the energy of the incident electrons.
From equation (1), the atomic spacing between (220) planes is
a/220 = [tex]\sqrt{2}[/tex] / 220 = 5.43 × 10⁻¹⁰ / [tex]\sqrt{2}[/tex]
Using equation (3), the wavelength of the incident electron beam is given as
kλ = 2d sinθ = 2 x 5.43 × 10⁻¹⁰ /[tex]\sqrt{2}[/tex] × sin 3.66°
= 1.631 × 10⁻¹⁰ m
Now using equation (2), the energy of the incident electrons is
E = (hc) / λ = (6.626 × 10⁻³⁴ Js) × (3 × 10⁸ m/s) / 1.631 × 10⁻¹⁰ m
= 1213 eV
= 1.213 keV
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Two point charges, Q1 and Q2, are located at (1, 2, 0) and (2, 0, 0), respectively. Find the relation between Q1 and Q2 such that the force on a test charge at the point P(-1, 1, 0) will have
a. No x-component
b. No y-component
the relation between Q1 and Q2 such that the force on the test charge at P will have no x-component is Q1 * sqrt(10) = Q2 * sqrt(13).
To find the relation between Q1 and Q2 such that the force on a test charge at the point P(-1, 1, 0) will have no x-component, we need to set up an equation using Coulomb's law. Coulomb's law states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's assume Q1 is the charge at (1, 2, 0) and Q2 is the charge at (2, 0, 0). The equation for the force on the test charge at P(-1, 1, 0) is:
F = k * (Q1 * Q2) / r^2
where k is the Coulomb's constant and r is the distance between the test charge and the point charge.
Now, since we want the force to have no x-component, we can set the x-components of the forces from Q1 and Q2 equal to each other. Let's denote the distance between Q1 and P as r1 and the distance between Q2 and P as r2.
The x-component of the force from Q1 is given by:
F1x = k * (Q1 * Q) / r1^2
The x-component of the force from Q2 is given by:
F2x = k * (Q2 * Q) / r2^2
Since we want F1x = F2x, we can equate the two equations:
k * (Q1 * Q) / r1^2 = k * (Q2 * Q) / r2^2
Canceling out the constants and rearranging the equation, we get:
(Q1 * Q) / r1^2 = (Q2 * Q) / r2^2
Now, substituting the values for r1, r2, and the coordinates of P, we have:
(Q1 * Q) / (sqrt(2^2 + 3^2)) = (Q2 * Q) / (sqrt(3^2 + 1^2))
Simplifying further:
(Q1 * Q) / sqrt(13) = (Q2 * Q) / sqrt(10)
Cross multiplying:
(Q1 * Q * sqrt(10)) = (Q2 * Q * sqrt(13))
Dividing both sides by Q * Q:
Q1 * sqrt(10) = Q2 * sqrt(13)
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A circular core of high relative permeability material is shown in figure Q2c. Insulated wire is wrapped around the core to make an inductor of 100mH. An alternating current source is connected in series with a resistor to the magnetic circuit. The supplied current is sinusoidal with a RMS (Root Mean Square) current of 10 mA at a frequency of 12kHz. i) Determine the RMS voltage across the inductor. The AC current source is replaced by a DC current source. The output is initially isolated from the magnetic circuit by an open switch. The DC supply is set to 10 mA. The switch is then closed and the current is allowed to flow.
RMS voltage across the inductor is determined using the formula for RMS voltage across an inductor, Vrms = 2πfLI where L is the inductance of the inductor and I is the RMS current supplied to the inductor. The frequency f is given to be 12 kHz, and the inductance L is given to be 100 mH.Explanation :The inductor has an inductance L = 100 mHThe frequency f is given to be 12 kHz
The RMS current supplied to the inductor is I = 10 mA.The formula for RMS voltage across an inductor is given as:Vrms = 2πfLIOn substituting the values of L, f and I in the formula, we get,Vrms = 2π(12 kHz)(100 mH)(10 mA) = 7.54 Vii) The magnetic circuit has a circular core of high relative permeability material and an insulated wire wrapped around the core to form an inductor. A DC current source is connected to the circuit. Initially, the output of the DC current source is isolated from the magnetic circuit by an open switch.
Then, the DC supply is set to 10 mA and the switch is closed. As the switch is closed, the current starts to flow in the circuit through the inductor. However, since the core is made up of a high permeability material, the inductance of the circuit increases. This leads to a delay in the buildup of current in the circuit. Due to the delay, the current in the circuit does not reach the steady-state immediately after the switch is closed.
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